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  • Need clarity, kindly explain! - Complex numbers and quadratic equations - JEE Main-6

Conjugate of  \frac{\left ( 3-i \right )\left ( 2+i \right )}{\left ( 1-i \right )\left ( 3+i \right )}  will be

  • Option 1)

    \frac{13}{10}+\frac{9}{10}i

  • Option 2)

    \frac{-13}{10}+\frac{9}{10}i

  • Option 3)

    \frac{13}{10}-\frac{9}{10}i

  • Option 4)

    \frac{-13}{10}-\frac{9}{10}i

 

Answers (1)

best_answer

 \frac{\left ( 3-i \right )\left ( 2+i \right )}{\left ( 1-i \right )\left ( 3+i \right )}=\frac{6-i^{2}+3i-2i}{3-i^{2}-3i+i}=\frac{7+i}{4-2i}=\frac{7+i}{4-2i}=\frac{26+18i}{20}=\frac{13+9i}{10}=\frac{13}{10}-\frac{9}{10}i

\therefore its conjugate will be \frac{13}{10}-\frac{9}{10}i

\therefore Option (C)

 

Conjugate of a Complex Number -

z=a+ib \Rightarrow \bar{z}=a-ib

- wherein

\bar{z} denotes conjugate of z

 

 


Option 1)

\frac{13}{10}+\frac{9}{10}i

This is incorrect

Option 2)

\frac{-13}{10}+\frac{9}{10}i

This is incorrect

Option 3)

\frac{13}{10}-\frac{9}{10}i

This is correct

Option 4)

\frac{-13}{10}-\frac{9}{10}i

This is incorrect

Posted by

Aadil

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